Timeline for Lp estimate for resolvent of Laplace operator
Current License: CC BY-SA 3.0
8 events
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| Aug 25, 2014 at 17:57 | comment | added | Christian Remling | You can of course solve $y''-zy=f$, $y'(0)=y'(1)=0$ explicitly. This gives a (simple) formula for $G$. As for the estimate being coarse, I'm sorry to learn that you don't like it. I was addressing the question you asked. | |
| Aug 25, 2014 at 17:23 | comment | added | Marcin Malogrosz | Which explicit formula do You have in mind? From your reasoning $||R(z,A_p)||_{\mathcal{L}(L_p)}\leq ||||G(x,t;z)||_{L_p(dx)}||_{L_{p'}(dt)}$. Now if $p=2$ one has $||||G(x,t;z)||_{L_p(dx)}||_{L_{p'}(dt)}=||G(x,t;z)||_{L_2(dx\otimes dt)}=(\sum_{n=0}^{\infty}1/|\lambda_n-z|^2)^{1/2}$, so the estimate seems to be a bit to coarse even for $p=2$. | |
| Aug 25, 2014 at 16:55 | comment | added | Christian Remling | If you want to analyze $|z|\to\infty$, I would try to work with the explicit formula for $G$ and use facts about Fourier transforms. | |
| Aug 25, 2014 at 16:54 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Aug 25, 2014 at 16:43 | comment | added | Christian Remling | You mean for $z\to\infty$? I didn't consider this case; I focused on $z$ between eigenvalues (as mentioned in your question). | |
| Aug 25, 2014 at 16:21 | comment | added | Marcin Malogrosz | For me it is not clear. If $z>0$ you need to show that $\sum_{n=0}^{\infty}\frac{z}{z+n^2}$ is bounded in $z$ which is clearly not. | |
| Aug 25, 2014 at 12:53 | vote | accept | Marcin Malogrosz | ||
| Aug 25, 2014 at 16:21 | |||||
| Aug 25, 2014 at 3:53 | history | answered | Christian Remling | CC BY-SA 3.0 |